Notesale: Turn your study into money

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Integration
An integral is an infinite sum of infinitesimal terms
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I want to convey what I mean by the term “operator” by first
discussing an operator with which you are already familiar, namely the derivative operator
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The
d
(the derivative with respect to x) acts on a function of x to yield another
derivative operator
dx
function of x
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We write the derivative of y(x) as
y or

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The integral operator is an operator too
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(But see footnote 1
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The point here, is not about the order
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) The point is that the integration
operator includes both the integral sign “

∫

” and the differential “dx”
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An infinite sum of finite terms is infinite
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So, given the function f (x) ,
∫ f (x) is nonsense, whereas, ∫ f ( x) dx is the integral of f with respect to x
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Consider the function f ( x ) = x 2
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The bottom line is, if you want to stick
an integral sign “ ∫ ” in front of something, you better make sure that something has a

whereas,

2

2

differential in it, otherwise, your final expression is nonsense
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Consider an object whose velocity as a function of time is given by v (t ) = 1
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s
Assume you need to know how far it goes during the first 4
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At first you
might be tempted to use your junior high school formula “distance is speed times time”
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0 seconds but what are you going to use for the speed
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0 seconds you get the speed at the 4 second mark but the object is not going that fast for the
whole four-second time interval from 0 to 4
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Its speed is 0 at time zero, 1
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0 m/s
at 2 seconds, 13
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The object is clearly speeding up
during the entire 4 seconds
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But that only works if the

1
2

I am using the generic variable name x
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It could be t or z or anything else
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Infinitesimal is vanishingly small
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1

m 2
t to find that the velocity is
s3
12 m/s at time t = 2
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So, in the case at hand, the object goes slower than 12 m/s for the first
2
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So the average speed has to be less
than 12 m/s
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5 3 t 2 , then the acceleration of the object is not constant
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So how do we go about finding out how far the object goes in the first 4
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Note that you can solve v (t ) = 1
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But, in general, there is
no easy way to get the average speed
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0 seconds?
Newton is the one that came up with the idea
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Here’s the idea
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Let’s try four parts at first
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The distance will be the average speed during the 1st second, times one
second; plus the average speed during the 2nd second, times one second; plus; the average speed
during the 3rd second, times one second; plus the average speed during the 4th second, times one
second
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Still, whether I use, as my estimated speed for the average speed
during the time interval, in each case, the speed at the start of the one-second time interval, the
speed at the end of the one second time interval, or the sum of the two divided by 2, the end
result will be closer to the actual value than I’d get by using the corresponding value for the
entire four-second time interval
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So, using any reasonable method to “guess”
the average speed during each one second time interval will yield a result closer to the actual
value than using the same method to arrive at a guess for the entire four second time interval
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But Newton was smarter than
that
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Rather, that we should use shorter time intervals
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Suppose we divide the
original four-second time interval up into four million equal time intervals
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For each millionth of a second, from zero to four seconds, if we multiply
the average speed during that millionth of a second, times one millionth of a second, and add all
the results together, we get the distance traveled during the first four seconds
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2

Check it out
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Every term in
our sum will be a little bit on the low side because the actual average speed during the millionth
of a second time interval will be a little greater than the speed at the start of the time interval
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This results in a
distance that is greater than the actual distance
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Your second sum of four
million terms will look just like the first one, but, it will be missing the first term (0 m/s times
0
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000001 s), that does not
appear in the first sum
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000024 m greater than the first attempt
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The actual distance is somewhere in between
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000024 m
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You’d think that would be good enough for Newton
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000001 s time intervals is already pretty extreme in my book
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He divided it up into an infinite number of infinitesimal time intervals
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So he winds up with an
infinite sum of infinitesimal elements, something we now call an integral
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0s

∆x = ∫ v (t ) dt
0

It represents the infinite sum of terms

v (0)dt +v (dt )dt +v (2dt )dt +v (3dt )dt +
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0 s, we evaluate v at t and
multiply the result by dt
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Then we increase t by dt
and repeat until t reaches the value of 4
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The end result of the sum is the distance traveled by
the object from time 0 to time 4
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4
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In the expression ∆ x = ∫ v (t ) dt , the variable t is referred to as the
0

variable of integration
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The 0 is the lower limit of
integration and the 4
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The equation reads, “Delta x is equal to
the integral of v (t) from zero to four seconds”
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The function is called the antiderivative of the function that is being
integrated
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0 s

∫ v (t ) dt = x(4
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Fortunately, all of the hard
work has been done for us my the mathematicians, so here, in your physics course, we can
provide you with a few simple rules for arriving at the results
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Here, we simply
present them to you, along with some information on notation and usage, without proof
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m
We again rely on the example of the object whose velocity is given by v (t ) = 1
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First off, as you know, the velocity v (t) is just the time derivative of the position
4
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0 s
dx
dx
variable x: v (t ) =

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We
dt
dt
0
0
4
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We don’t need any fancy rules

0

to interpret this
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0s
...
0s) – x(0)
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(But see footnote 3
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Start with x
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Now integrate that
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Integration is the inverse operation
dt
to taking the derivative
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If you integrate a function, and then take the derivative of the result, you get the original function
back
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Here, when we say “integrate a function” we really
mean “find the antiderivative of a function
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(As you might guess, an integral that includes the limits
of integration is called a definite integral
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4
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5

0
4
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5 s

3

m 2
t
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Now this represents an infinite sum in which every term is being

0

multiplied by the constant 1
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We can factor that constant out of the sum and write the
s3

3

The position function is actually the initial position plus the antiderivative
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In general, when you find an antiderivative of f (x) you are finding a function g(x) whose
derivative is f (x)
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Call the result h(x) = g(x) + constant
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So if the derivative of g(x) is f (x) then the derivative of h(x) = g(x) + constant is also f (x)
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4

integral as 1
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0 s

2
∫ t dt
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0 s

∫t

0

2

dt part we need to find a function whose

0

2

derivative is t
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The derivative of t 3 is 3t 2
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That makes t our current candidate for the antiderivative of t
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3
1
2
Now we introduce a bit more notation
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0s
4
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5 3 ∫ t 2 dt = 1
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0 s

That 0 part is to be read “evaluated from 0 to 4
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So the next line reads
1
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0s

∫t

2

dt = 1
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5 3
s

m1
1

( 4
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0 s

∫t

2

dt = 32 m

0

4
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In the process we have
s3
come up with a rule for the integral of a power
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5

Let’s write that power rule as a function of x
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An antiderivative of xn can be arrived at by
incrementing the power by 1 and dividing by the new power

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